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31151.

The de Broglie wavelength of electron in second Bohr orbit is exactly equal to

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the CIRCUMFERENCE of the orbit
Double the circumference of the orbit
Half the circumference of the orbit
thrice the circumference of the orbit

Solution :ANGULAR MOMENTUM `mvr=(nh)/(2PI)`
`mv=(2h)/(2pir)=h/(PIR)`
And `lamda=h/(mv)=(hxxpir)/h=pir`
i.e. half the circumference.
31152.

The de Broglie wavelength of an electron is 66 nm. The velocity of the electron is (h=6.6xx10^(-34)kgm^(2)s^(-1)m=9.0xx10^(-31)kg)

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`1.84xx10^(-14)ms^(-1)`
`1.1xx10^(-4)ms^(-1)`
`5.4 XX 10^(3)ms^(-1)`
`1.1xx10^(4)ms^(-1)`

Solution :`lamda=h/(mv)=(6.6xx10^(-34))/(9.0xx10^(-31)xx66xx10^(-9))`
`=1.1xx10^(4)ms^(-1)`
31153.

The de Broglie wavelength of a particle with mass 1 g and velocity 100 m/s is:

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`6.63xx10^(-33)`m
`6.63xx10^(-34)`m
`6.63xx10^(-35)`m
`6.65xx10^(-35)`m

Answer :A
31154.

The de Broglie wavelength associated with a particle of mass 10^-6 kg with a velocity of 10 ms^-1 is (h=6.625 × 10^-34Js)

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`6.626×10^-34`m
`6.626×10^-29`m
`6.626 × 10^-28` m
`6.626 × 10^-40`m

Answer :B
31155.

The de Broglie wavelength associated with a material particle is:

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INVERSELY PROPORTIONAL to MOMENTUM
Inversely proportional to its energy
DIRECTLY proportional to momentum
Directly proportional to its energy

Answer :A
31156.

The dcecomposition of a substance follows first order kinatics. Its concentraction is reduced to (1)/(8)th of its initial value in 24 minutes. The rate constant of the decomposition process is

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`(1)/(4)mi N^(-1)`
`(0.692)/(24)MOL^(-1)`
`(2.303)/(24)"LOG"((1)/(8))mi n^(-1)`
`(2.303)/(24)"log"((8)/(1)mi n^(-1)`

SOLUTION :`k=(2.303)/(t)"log"(a)/(a-x)=(2.303)/(24)"log"((1)/((1)/(8)))=(2.303)/(24)"log"8`
31157.

The data given below are for the reactionof NO and Cl_(2) to form NOCl at 295 K [Cl_(2)]""[NO]"""initial rate(mol litre"^(-1)"sec"^(-1)) 0.05""0.05""1xx10^(-3) 0.15""0.05"3xx10^(-3) 0.05""0.15""9xx10^(-3) If [Cl_(2)] is halved and (NO) is double the rate of the reaction is ________

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halved
doubled
unchanged
becomes 1/4th

Solution :`(r_(2))/(r_(1))=(KXX(2)^(2)xx(1/2)^(1))/(Kxx(1)^(2)xx(1)^(1)),(r_(2))/(r_(1))=4xx1/2=2impliesr_(2)=2xxr_(1)`
31158.

The data given below are for the reactionof NO and Cl_(2) to form NOCl at 295 K [Cl_(2)]""[NO]"""initial rate(mol litre"^(-1)"sec"^(-1)) 0.05""0.05""1xx10^(-3) 0.15""0.05"3xx10^(-3) 0.05""0.15""9xx10^(-3) The reaction rate when conc. of Cl_(2) an NO are 0.2 M and 0.4 M respectively is

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0.256 MOLE `"litre"^(-1)`
0.256 mole `"litre"^(-1)"sec"^(-1)`
0.512 mole litre `"sec"^(-1)`
0.512 mole `"litre"^(-1)`sec

Solution :`r=K[NO]^(2)[Cl_(2)]^(1)=8XX(0.4)^(2)XX(0.2)^(1)=8xx0.4xx0.4xx0.2=0.256`
31159.

The data given below are for the reaction of NO and Cl_(2) to form NOCl at 295 K [Cl_(2)]""[NO]"""initial rate(mol litre"^(-1)"sec"^(-1)) 0.05""0.05""1xx10^(-3) 0.15""0.05"3x10^(-3) 0.05""0.15""9xx10^(-3) The rate constant of the reaction is

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6
8
3
2

Solution :`NO+1/2Cl_(2)toNOCl,r=K[NO]^(x)[Cl_(2)]^(y)`
`1xx10^(-3)=K(0.05)^(x).(0.05)^(y)`……1, `3xx10^(-3)=K(0.05)^(x).(0.15)^(y)`………2
`9xx10^(3)=K.(0.15)^(x).(0.05)^(y)`……….3, `((3))/((1))implies((9xx10^(-3))/(1xx10^(-3)))=(Kxx(0.15)^(x)XX(0.05)^(y))/(Kxx(0.05)(0.05))`
`3^(2)=3^(x)impliesx=2,((2))/((1))implies((3xx10^(-3))/(1xx10^(-3)))=(Kxx(0.05)^(x)xx(0.15)^(y))/(Kxx(0.05)(0.05))`
`3^(1)=3^(y)impliesy=1,r=K.[NO]^(2)[Cl_(2)]^(1),K=8`
31160.

The data for the conversion of compound A into its isomeride B are as follows : {:("Time is hr :",,,0,,,1,,,2,,,3,,,4,,,oo),(%"age of A :",,,49.3,,,34.6,,,25.8,,,18.5,,,13.8,,,4.8):} Show that this is a first order reaction.

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Solution :`aprop49.3,(a-x)prop35.6" at "t=1" hr and so on. Substitute in 1ST ORDER eqn. k COMES out to be constant."`
31161.

The dark blue colour of the solution formed when excess of ammonia is added to a solution of copper (II) sulphate is due to the presence of the ion

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`[Cu(OH)_(4)(H_(2)O)_(2)]^(2-)`
`[Cu(H_(2)O)_(6)]^(2+)`
`[Cu(NH_(3))_(2)(H_(2)O)_(4)]^(2+)`
`[Cu(NH_(3))_(4)(H_(2)O)_(2)]^(2+)`

ANSWER :D
31162.

The d_(111)spacing for crystalline K is 0.3079 ran, Calculate the length of the cubic unit cell.

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ANSWER :0.5333 NM
31163.

The 'd' orbitals will be split under square planar geometry into

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two levels
THREE levels
FOUR levels
five levels

ANSWER :C
31164.

The d-orbitals participating in hybridisation of metal atom may be from the n & (n-1) shell. This depends on the nature of metal and nature of ligand. The complexes involving the inner d-level (inner orbital complexes) results when the ligand is a powerful (or) strong ligand results in a diamagnetic (or) low spin complexes. A weak ligand usually results in the formation of outer orbital complex (or) high spin complex. Among the following sets, the one having the same geometry and same magnetic property for both complexes is

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`[Mn(CN)_(6)]^(3-)&[MnF_(e)]^(3-)`
`[CO(CN)_(6)]^(3-)&[CoF_(6)]^(3-)`
`[Fe(CN)_(6)]^(4-)&[Fe(H_(2)O)_(6)]^(2+)`
`[CR(CN)_(6)]^(3-)&[Cr(NH_(3))_(6)]`

Solution :`[Cr(CN)_(6)]^(3-),d^(2)sp^(3)` OCTAHEDRAL.
`[Cr(NH_(3))_(6)]:d^(2)sp^(3)` octabedral.
`:.` Both `CN^(-)` and `NH_(3)` are strong field ligands.
31165.

The d-orbitals participating in hybridisation of metal atom may be from the n & (n-1) shell. This depends on the nature of metal and nature of ligand. The complexes involving the inner d-level (inner orbital complexes) results when the ligand is a powerful (or) strong ligand results in a diamagnetic (or) low spin complexes. A weak ligand usually results in the formation of outer orbital complex (or) high spin complex. Number of unpaired electrons present in [Fe(CN)_(6)]^(4-)&[Fe(H_(2)O)_(6)]^(2+) are respectively.

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4,4
0,0
0,4
2,4

Solution :`CN^(-)` : Strong field ligand and electronic rearrangment is POSSIBLE `H_(2)O` weak field ligand and no ELECTRIC rearrangement.

(in presence of `H_(2)O)` )

(in presence of `CN^(-)` )
31166.

The d-orbitals participating in hybridisation of metal atom may be from the n & (n-1) shell. This depends on the nature of metal and nature of ligand. The complexes involving the inner d-level (inner orbital complexes) results when the ligand is a powerful (or) strong ligand results in a diamagnetic (or) low spin complexes. A weak ligand usually results in the formation of outer orbital complex (or) high spin complex. The hybridisation of Cu in [CuCl_(4)]^(2-)&[Cu(NH_(3))_(4)]^(2+) are respectivly

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`DSP^(2),dsp^(2)`
`sp^(3),dsp^(2)`
`sp^(3),sp^(3)`
`dsp^(2),sp^(3)`

Solution :Cu UNDERGOES `sp^(3)` hybridization in `[CuCl_(4)]^(2-)`

In `[Cu(CN)_(4)]^(2-)`, rearrangement of UNPAIRED `e^(-)` TAKES PLACE as shown and Cu undergoes `dsp^(2)` hybridisation.
31167.

The d-orbitals involved in sp^(3)d^(2) or d^(2)sp^(3) hybridization of the central metal ion are:

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`d_(x^(2)-y^(2))`
`d_(XY)`
`d_(YZ)`
`dx_(z^(2))`

Answer :A::D
31168.

The d-orbitals involved in sp^(3)d^(2) or d^(2)sp^(3) hybridisation of the central metal ion are :

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`d_(X^(2)-y^(2)) & d_(z^(2))`
`d_(xy) & d_(z^(2))`
`d_(YZ) & d_(x^(2)-y^(2))`
`d_(z^(2)` only

Solution :Information BASED
31169.

The d orbitals involved in d^(2)sp^(3) hybridization are …………….

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`d_(xy)`, `d_(YZ)`
`d_(x^(2)-y^(2))`, `d_(z^(2))`
`d_(ZY)`, `d_(XZ)`
`d_(xy)`, `d_(z^(2))`

ANSWER :B::D
31170.

The d-orbitals involved in sp^(2)d and sp^(3)d^(2) hybridisation are respectively :

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`d_(X^(2) - y^(2)) ` and `d_(x^(2) -y^(2)),d_(x^(2))`
`d_(Z^(2))` and `d_(XY) ,d_(yz)`
`d_(x^(2)-y^(2))` and `d_(xy),d_(zx)`
`d_(x^(2)) ` and `d_(x^(2)-y^(2)) ,d_(x^(2))`

Answer :A
31171.

The d orbital involved in the dsp^(3) hybridisation of Fe(CO)_(5)] is ...............

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ANSWER :`d_(Z^(2))`
31172.

The d orbital involved in dSP3 hybridisation of [Fe(CO)_(5)] is………..

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`d_(XY)`
`d_(yz)`
`d_(xz)`
`d_(X^(2)-y^(2))`

Answer :B::D
31173.

The d-electron configuration of Mn^(2+) and Ti^(2+) is d^(5) and d^(2), respectively. Which one of these ions will be more paramagnetic ?

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Solution :`Mn^(2+)` has 5 UNPAIRED electrons WHEREAS `TI^(2+)` has 2 unpaired electrons. Therefore, `Mn^(2+)` is more PARAMAGNETIC than `Ti^(2+)`.
31174.

The d-electron configuration of Cr^(2+), Mn^(2+),Fe^(2+) and Co^(2+)are d^4, d^5,d^6 and d^7 respectively. Which one of the following will exhibit minimum paramagnetic behaviour ? (At. Nos. Cr=24 , Mn=25 , Fe=26 , Co=27)

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`[CR(H_2O)_6]^(2+)`
`[MN(H_2O)_6]^(2+)`
`[FE(H_2O)_6]^(2+)`
`[CO(H_2O)_6]^(2-)`

ANSWER :D
31175.

The d-block elements form______________compounds and complexes.

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SOLUTION :INTERSTITIAL
31176.

The d-block consisting of groups______________occupies the large middle section of the periodic table.

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SOLUTION :`3 - 12`
31177.

The d and l enantiomers of an optically active compound differ in

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Their boiling and melting point
Their rotation of plane polarized light
Their solubility
Their REFRACTIVE index

Solution :The d and L enantiomers of an OPTICALLY active compound DIFFER in their rotation of plane polarised light. The d enantiomer rotates it to the right whereas the l rotates it to the left.
31178.

The cyclobutyl methylamine with nitrous acid gives

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All of these

Answer :D
31179.

The cyclic form of fructose is called……………

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SOLUTION :FURANOSE FORM
31180.

The cyclic structure of glucose with 5 carbon and one oxygen atom is called……………

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SOLUTION :pyranoseform
31181.

The cyclic ether tetrahydrofuran (THF) can be synthesized by treating 4-chloro-1-butanol with aqueous sodium hydroxide (see below). Propose a mechanism for this reaction.

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Solution :Removal of a proton from the HYDROXYL group of 4-chloro 1-butanol gives an alkoxide ion that can then REACT with itself in an intramolecular `S_(N)2` reaction to form a ring.

EVEN though treatment of the alcohol with hydrox ide does not favor a large equilibrium concentration of the alkoxide, the alkoxide anions that are present react rapidly by the intramolecular `S_(N)2`reaction. As alkoxide anions are consumed by the substitution reaction, their equilibrium concentration is REPLENISHED by deprotonation of additional alcohol molecules, and the reaction is drawn to completion.
31182.

The cyanide ion, CN^(-), N_(2) and N_(2) are isoelectronic. But in contrast to CN^(-), N_(2) is chemically inert, because of

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Low bond energy
Absence of bond POLARITY
Unsymmetrical electron distribution
Presence of more number of ELECTRONS in bonding ORBITALS

Solution :SIMILAR atoms are linked to each other and thus there is not polarity.
31183.

The curves obtained when molar conductivity lambda_(m) (along Y-axis) is plotted against the square root of concentration C^(1//2) (along X-axis) for two electrolytes 'A' and 'B' are shown. (a) What can you say about the nature of the two electrolytes ? (b) How do you account for the increase in molar conductivity Lambda_(m) for the electrolytes A and B on dilution ?

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Solution :(a) ELECTROLYTE 'A' is a strong while 'B' is a weak electrolyte.
(b) In case of strong electrolyte, 'A', the ionisation is already COMPLETE. With dilution, the interionic forces decrease. Therefore, the ionic mobility increases leading to small increase in the value of `Lambda_(m)`. This continues till LIMITING value of molar conductance reaches. It is denoted either as `Lambda^(0)` or as `Lambda^(oo)`. In case of weak electrolyte, 'B' the EXTENT of dissociation is very small. It increase with dilution. As a result, the value of `Lambda_(m)` increases with dilution. Since there is an enormous increase in ionisation upon dilution, there is large increase in the value of `Lambda_(m)` when the solution is very dilute. This has been depicted in the curve. In case of electrolyte 'B', the limiting value of the molar conductance (`Lambda_(m)^(0)`) cannot be achieved since the dissociation of the electrolyte is never complete.
31184.

The cyanide ion, CN^(-) and N_(2) are isoelectronic. But in contrast to CN^(-), N_(2) is chemically inert because of

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low bond energy
absence of bond polarity
unsymmetrical ELECTRON distribution
presence of more NUMBER of electrons in bonding ORBITALS.

Answer :2
31185.

The cyanide ion, CN^(-) and N_(2) are isoelectronic. But in contrast to CN^(-),N_(2) is chemically inert, because of

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low bond energy
absence of bond polarity
unsymmetrical electron distribution
presence of more number of electrons in bonding ORBITALS.

Solution :Both `CN^(-)` and `N_(2)` MOLECULES have the same MO DIAGRAMS but differ in bond polarity. Thus, `N_(2)` is inert DUE to absence of bond polarity while `CN^(-)` ion is reactive due to presence of bond polarity.
31186.

The cyanide complex of silver formed in the silver extraction in Mac-Arthur's Forrest cyanide process is:

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`[Ag(CN)_(2)]^(-)`
`K_(2)[Ag(CN)_(3)]`
`[Ag(CN)_(4)]^(2-)`
`Na_(2)[Ag(CN)_(4)]`

ANSWER :A
31187.

The value of enthalpy change (Delta H) for the reaction : C_(2)H_(5)OH(l) + 3O_(2)(g) rarr 2CO_(2)(g) + 3H_(2)O(l) at 27^(@)C is - 1366.5 kJ mol^(-1). The value of internal energy change for the above reaction at the temperature will be :

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`- 1369.0 kJ`
`- 1364.0 kJ`
`- 1361.5 kJ`
`- 1371.5 kJ`

Solution :`DELTA H = Delta U + Delta n_(g) RT`
or `Delta U = Delta H + Delta n_(g) RT`
`Delta n_(g) = 2 - 3 = -1`
:. `Delta U = - 1366.5 - (- 1) xx 8.3 xx 10^(-3) xx 300`
`= - 1366.5 + 2.490 = - 1364.01 kJ`
31188.

The curve showing variation of x/m with temperature is inverted 'V' shape. What is the type of adsorption ?

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SOLUTION :CHEMISORPTION
31189.

The curve showing the variation of pressure with temperature for a given amount of adsorption is called

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ADSORPTION isobar
adsorption isotherm
adsorption isostere
adsorption isochore.

Solution :The curve showing the VARIATION of PRESSURE with temperature is CALLED adsorption isostere.
31190.

The curve showing the variation of pressure with temperature for a given amount of adsoption is called

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ADSORPTION isobar
adsorption isotherm
adsorption isotere
adsorption isochore

Answer :D
31191.

The curve showing the variation of adsorption with pressure at constant temperature is is called :

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An isostere
ADSORPTION isobar
Adsorption isotherm
All

Answer :B
31192.

The curve showing the variation of adsorption with pressure at constant temperature is called:

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ADSORPTION ISOTHERM
Adsorption ISOBAR
An Isostere
Absorption isobar.

Answer :A
31193.

The curve represents the titration of :

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a diprotic acid
TWO MONOPROTIC acids with the same `K_(a^(s))` but different CONCENTRATIONS.
two monoprotic acid with different `K_(a^(s))` but the same concentrations
two monoprotic acids with different `K_(a^(s))` and different concentrations.

ANSWER :D
31194.

The curve of pressure volume (PV) against pressure (P) of the gas at a particular temperature is as shown, according to the graph which of the following is incorrect (in the low pressure region):

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`H_(2)` and He shows +ve deviation from ideal GAS equation.
`CO, CH_(4)` and `O_(2)` show negative deviation from ideal gas equation.
`H_(2)` and He show negative deviation while `CO, CH_(4)` and `O_(2)` show positive deviation.
`H_(2)` and He are LESS compressible than that of an ideal gas while `CO, CH_(4)` and `O_(2)` more compressible than that of ideal gas.

Solution :If `Z gt 1` positive deviation
`Z LT 1` negative deviation
31195.

The curve of pressure volume (PV) against pressure (P) of the gas at a particular temperature is as shown, according to the graph which of the following is/are incorrect (in low pressure region):

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`H_2` and He show POSITIVE deviation from ideal gas EQUATION
CO, `CH_4` and `O_2` show NEGATIVE deviation from ideal gas equation
`H_2` and He show negative deviation while `CO_2,CH_4` and `O_2` show positive deviation
`H_2` and He are less compressible than that of an ideal gas while `CO_2, CH_4` and `O_2` more compressible than that of ideal gas.

Solution :if `ZGT1` positive deviation
`Zlt1` negative deviation
31196.

The curve of pressure volume (PV) against pressure (P) of the gas at a particular temperature is as shown, according to the graph. Which of the following is incorrect (in the low pressure region)?

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`H_(2)` and He SHOWS +ve deviation from ideal gas equation
`CO, CH_(4)` and `O_(2)` show negative deviation from ideal gas equation.
`H_(2)` and He show negative deviation while `CO_(2), CH_(4)` and `O_(2)` show POSITIVE deviation
`H_(2)` and He are LESS compressible than that of an ideal gas while `CO_(2),CH_(4)` and `O_(2)` more compressible than that of ideal gas.

Answer :C
31197.

The curve in the accompanying diagram represent the PV//RT behaviour of the gases: He, CH_(4) and C_(3)H_(8). Which assignment of behavior for gas is correct ?

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`1 = He, 2 = CH_(4), 3 = C_(3)H_(8)`
`1 = C_(3)H_(8), 2 = CH_(4), 3 = He`
`1 = CH_(3), 1 = C_(3)H_(8), 3 = He`
`1 = C_(3)H_(8), 2 = He, 3 = CH_(4)`

Answer :B
31198.

the current statement on the Aufbau principle is that

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(n-1)d subshell is always lower in ENERGY than NS orbital
(n-1)F subshell always has energy more thannp subshell
5d is lower in energy than 4f
6p is lower in energy than 5d

Answer :B
31199.

The current strength required to displace 0.1 g. of H_2 in 10 sec is

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9.65 AMP
1.988 amp
198 amp
965 amp

ANSWER :D
31200.

The current required to diplace 0.1 g of H_(2) in 10 seconds will be

Answer»

9.65 amp
1.988 amp
198 amp
965 amp

Answer :D